Solution
Figure 1: Velocity vectors in the rocket propulsion system.
1. Defining Power Terms
Efficiency of propulsion ($\eta$) is defined as the ratio of useful power output to the total power input (rate of energy expenditure). $$ \eta = \frac{P_{useful}}{P_{input}} $$
- Useful Power ($P_{useful}$): This is the power generated by the thrust force propelling the rocket. $$ F_{thrust} = u \frac{dm}{dt} $$ $$ P_{useful} = F_{thrust} \cdot v = \left( u \frac{dm}{dt} \right) v $$
- Wasted Power ($P_{waste}$): The problem states that the kinetic energy imparted to the burnt fuel is wasted. The absolute velocity of the ejected gas (with respect to the ground) is $v_{gas} = |v – u|$. $$ P_{waste} = \frac{1}{2} \frac{dm}{dt} (v – u)^2 $$
2. Energy Balance
The total input power comes from the chemical energy of the fuel, which is converted into the useful work on the rocket plus the wasted kinetic energy of the exhaust. $$ P_{input} = P_{useful} + P_{waste} $$ $$ P_{input} = u \dot{m} v + \frac{1}{2} \dot{m} (v – u)^2 $$
Expanding the term:
$$ P_{input} = \dot{m} \left[ uv + \frac{1}{2}(v^2 + u^2 – 2uv) \right] $$ $$ P_{input} = \dot{m} \left[ uv + \frac{1}{2}v^2 + \frac{1}{2}u^2 – uv \right] $$ $$ P_{input} = \frac{1}{2} \dot{m} (u^2 + v^2) $$3. Calculating Efficiency
Now we substitute $P_{useful}$ and $P_{input}$ into the efficiency formula: $$ \eta = \frac{P_{useful}}{P_{input}} = \frac{u \dot{m} v}{\frac{1}{2} \dot{m} (u^2 + v^2)} $$
Canceling $\dot{m}$:
